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3 years ago

Can someone please help me with this I’m sincerely confused?

Physics

2 answers:

Burka [1]3 years ago

6 0

Answer: im sure its the 4th one

Explanation:

77julia77 [94]3 years ago

4 0

Answer:

the fourth one

Explanation:

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What is the difference in intensity for two sounds that are 40 dB apart?

Galina-37 [17]

the difference of 10 dB is considered as having 10 times the intensity, so difference of 40 dB means :

=》10 × 10 × 10 × 10

=》10000 times

so the the intensity of one sound is 10000 times the other one.

4 0

3 years ago

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A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The r

ehidna [41]

Answer:

Mass of the planet = 6.0 × $10^{24}$

Explanation:

Time period = 2π (R + h) / v

Orbital speed (v) = √GM / (R + h)

T² = 4π² (R + h)² / (GM/ (R + h))

= 4π² (R + h)³ / GM

making m the subject of the formula

m = 4π² (R + h)³ / GT²

= 4π² ( 6.38 × $10^{6}$ + 230 × 10³ )³ / ( 6.67 × $10^{-11}$ ) × (89 × 60)²

= 4π² ( 6610000)³ / ( 6.67 × $10^{-11}$ ) × (89 × 60)²

= 5.99 × $10^{24}$

= 6.0 × $10^{24}$

5 0

3 years ago

An object has an initial velocity v and uniform acceleration a. If it covers a distance d, then its final velocity, v, is given

viktelen [127]

Answer:

Explanation:

5 0

3 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

6 0

3 years ago

3 things microwaves do

larisa86 [58]

Your Answer Is...

Heat up food, defrost food, make popcorn.

Glad I Could Help, And Good Luck!

~AnonymousGiantsFan~

5 0

3 years ago

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